Wrong-way Frobenius reciprocity for finite groups representations This is a typical lazy mathematician question, so do not hesitate to close it and recommend me to do my homeworks... 
Let $H$ be a subgroup of a finite group $G$, and let $Res_H^G$ and $Ind_H^G$ the restriction and induction functors between the categories of linear representations of $G$ and $H$, respectively. I'm familiar with Frobenius reciprocity stating that $Ind_H^G$ is the right adjoint to $Res_H^G$. Moreover, by choosing the traditional model for the induced representation, namely,
$Ind_H^G(U)=  \{f:G\to U \text{ such that } f(h g)=h\cdot f(g)\text{ for any } h\in H\}$
I also know how to write explicitly the natural isomorphism
$
Hom_{Rep(H)}(Res_H^G(W),U) \stackrel{\sim}{\to} Hom_{Rep(G)}(W,Ind_H^G(U)).
$
Now, I've just learnt that actually $(Res_H^G,Ind_H^G)$ is an ambidextrous adjunction, i.e. that $Ind_H^G$ is also the left adjoint to $Res_H^G$, but I'm not familiar with this fact: how is the isomorphism
$
Hom_{Rep(H)}(U,Res_H^G(W)) \stackrel{\sim}{\to} Hom_{Rep(G)}(Ind_H^G(U),W)
$
explicitly defined?
 A: The point is that there are two ways to describe the restriction functor. The first way is as $\text{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], -)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[G], \mathbb{C}[H])$-bimodule, which by the tensor-hom adjunction means that it has left adjoint $\mathbb{C}[G] \otimes_{\mathbb{C}[H]} (-)$. The second way is as $\mathbb{C}[G] \otimes_{\mathbb{C}[G]} (-)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[H], \mathbb{C}[G]$)-bimodule, which by a second application of the tensor-hom adjunction means means that it has right adjoint $\text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], -)$. 
It remains to write down a natural isomorphism between the left and right adjoints. Very explicitly, the obvious candidate is to try something like
$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_g) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$
where $V$ is a left $\mathbb{C}[H]$-module and $v_g \in V$. But we need to check that this is well-defined. On the LHS we have $gh \otimes v_{gh} = g \otimes h v_{gh} = g \otimes v_g$ for any $h \in H$, hence $h^{-1} v_g = v_{gh}$, whereas on the RHS we need functions such that if $g \mapsto v_g$ then $hg \mapsto hv_g = v_{hg}$ for any $h \in H$, hence $h^{-1} v_g = v_{h^{-1} g}$. These don't quite match up, so our map is slightly wrong. It should be
$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_{g^{-1}}) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$
and then everything matches up; the inverse map sends a function $(g \mapsto v_g)$ to $\sum g^{-1} \otimes v_g$.

Alternatively, instead of working with representations we can work with unitary representations on Hilbert spaces (in the finite-dimensional case this recovers the same theory). Here for every pair of unitary representations $V, W$ we have a natural identification $\text{Hom}(V, W) \cong \text{Hom}(W, V)$ given by taking adjoints, so categories of unitary representations are dagger categories. Any adjunction between dagger categories is automatically ambidextrous. 
In general, if $f : R \to S$ is a homomorphism of rings, we get a restriction functor $S\text{-Mod} \to R\text{-Mod}$ which can be described in either of the ways above and hence which has both a left and a right adjoint, which one might call "induction" and "coinduction" or something like that. These are left and right Kan extension along $f$ (thinking of $R, S$ as $\text{Ab}$-enriched categories with one object). 
